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Php 动态下拉菜单(如果有)_Php_Html_Drop Down Menu - Fatal编程技术网

Php 动态下拉菜单(如果有)

Php 动态下拉菜单(如果有),php,html,drop-down-menu,Php,Html,Drop Down Menu,我目前正在制作一对下拉菜单。第二个选项取决于用户在第一个下拉列表中的选择 目前我的代码不起作用,也没有进入if-else函数 $city=""; $formcontent =<<<EOT <select name="City" id="City"> <option value="0">Make a Selection</option> <option value=

我目前正在制作一对下拉菜单。第二个选项取决于用户在第一个下拉列表中的选择

目前我的代码不起作用,也没有进入if-else函数

$city="";
$formcontent =<<<EOT

            <select name="City" id="City">
            <option value="0">Make a Selection</option>
            <option value="City1" $city="City1">City1</option>
            <option value="City2" $city="City2">City2</option>
            </select><br></br>

            <select name="Street" id="Street">
            <option value="0">Make a Selection</option>
EOT;

                    if($city=="City1"){ 

$formcontent. = <<<EOT
                <option value="City1Street1">City1 Street1</option>
                <option value="City1Street2">City1 Street2</option>
EOT;
}
$city=”“;
$formcontent=您在附近有错误

if($city=="City1"){ 

$formcontent. = <<<EOT ...
if($city==“City1”){

$formcontent.=就在这里:要连接到变量,它是
$var.=
,而不是
$var.=
。修复:我更喜欢ajax根据所选城市给出街道列表。您只需将onchange事件应用到城市下拉列表,并根据传递的值,只需将innerhtml设置为街道下拉列表。if/else函数在哪里ion?是Javascript还是PHP?它仍然不会进入if语句。如果我选择city1,街道的选项中不会附加任何内容。如果要根据城市更改街道下拉选择,则需要使用Javascript/jquery。可以使用$(function(){$('#city').change(function(){var val=$(this).val();if(val='city1'){$('#street').html('street1street2');}否则if(val='city2'){$('#street').html('street3street4');}});
if($city=="City1"){ 

$formcontent .= <<<EOT ..